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# M22-36

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Math Expert
Joined: 02 Sep 2009
Posts: 43898

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16 Sep 2014, 00:17
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Difficulty:

15% (low)

Question Stats:

85% (00:56) correct 15% (01:44) wrong based on 66 sessions

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Circles $$X$$ and $$Y$$ are concentric. If the radius of circle $$X$$ is three times that of circle $$Y$$, what is the probability that a point selected inside circle $$X$$ at random will be outside circle $$Y$$?

A. $$\frac{1}{3}$$
B. $$\frac{\pi}{3}$$
C. $$\frac{\pi}{2}$$
D. $$\frac{5}{6}$$
E. $$\frac{8}{9}$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 43898

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16 Sep 2014, 00:17
Official Solution:

Circles $$X$$ and $$Y$$ are concentric. If the radius of circle $$X$$ is three times that of circle $$Y$$, what is the probability that a point selected inside circle $$X$$ at random will be outside circle $$Y$$?

A. $$\frac{1}{3}$$
B. $$\frac{\pi}{3}$$
C. $$\frac{\pi}{2}$$
D. $$\frac{5}{6}$$
E. $$\frac{8}{9}$$

We have to find the ratio of the area of the ring around the small circle to the area of the big circle. If $$y$$ is the radius of the smaller circle, then the area of the bigger circle is $$\pi(3y)^2 = 9 \pi y^2$$. The area of the ring $$= \pi(3y)^2 - \pi(y)^2 = 8 \pi y^2$$. The ratio $$= \frac{8}{9}$$.

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Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 432
Location: Netherlands
GMAT Date: 02-11-2015
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21 Jan 2015, 05:49
1
KUDOS
Hey,

Great that we saw how you do it using the actual variables.
However, I just used values.

For the radius of X = 6
For the radius of Y = 2

Then the area for X = 36π
and the area for Y = 4π

32/36 = 8/9.
Intern
Joined: 02 Aug 2017
Posts: 6
GMAT 1: 710 Q46 V41
GMAT 2: 600 Q39 V33

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26 Oct 2017, 14:20
I think the easiest way for me was:
$$\frac{πr^2}{π3r^2}$$

Use values Y=1, X=3Y=3

$$1^2 = 1, 3^2=9$$, 1/9 chance it is inside the circle, or 8/9 chance it is outside.
Re: M22-36   [#permalink] 26 Oct 2017, 14:20
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# M22-36

Moderators: chetan2u, Bunuel

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